Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{-2y + 6}{-3y^2 - 27y} \times \dfrac{y^3 + 8y^2 - 9y}{y^2 - 6y + 5} $
Solution: First factor out any common factors. $z = \dfrac{-2(y - 3)}{-3y(y + 9)} \times \dfrac{y(y^2 + 8y - 9)}{y^2 - 6y + 5} $ Then factor the quadratic expressions. $z = \dfrac {-2(y - 3)} {-3y(y + 9)} \times \dfrac {y(y - 1)(y + 9)} {(y - 1)(y - 5)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-2(y - 3) \times y(y - 1)(y + 9) } {-3y(y + 9) \times (y - 1)(y - 5) } $ $z = \dfrac {-2y(y - 1)(y + 9)(y - 3)} {-3y(y - 1)(y - 5)(y + 9)} $ Notice that $(y - 1)$ and $(y + 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-2y\cancel{(y - 1)}(y + 9)(y - 3)} {-3y\cancel{(y - 1)}(y - 5)(y + 9)} $ We are dividing by $y - 1$ , so $y - 1 \neq 0$ Therefore, $y \neq 1$ $z = \dfrac {-2y\cancel{(y - 1)}\cancel{(y + 9)}(y - 3)} {-3y\cancel{(y - 1)}(y - 5)\cancel{(y + 9)}} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $z = \dfrac {-2y(y - 3)} {-3y(y - 5)} $ $ z = \dfrac{2(y - 3)}{3(y - 5)}; y \neq 1; y \neq -9 $